Set 37 — Salient-Pole Alternators and Two-Reaction Theory
A cylindrical rotor presents the same air gap in every direction, so one synchronous reactance describes it completely. A salient-pole rotor does not: the gap under a pole face is short and the gap between poles is long, so flux produced by the armature meets a reluctance that depends on where the armature m.m.f. happens to point. Two reactances are needed, \(X_d\) along the pole axis and \(X_q\) between the poles.
The consequence is one extra step. Before the armature current can be given its reactance drops it must be split into a direct-axis part and a quadrature-axis part, and that split cannot be made until the position of the quadrature axis is known. Locating it is what the phasor \(\mathbf{V} + (R_a + jX_q)\mathbf{I}_a\) is for, and every problem in this set turns on that construction.
A salient-pole machine has two air gaps, not one. Flux entering under a pole face crosses a short gap; flux entering between the poles crosses a long one. The reluctance therefore depends on direction, and a single synchronous reactance can no longer describe the machine:
\[ X_d > X_q, \qquad \text{typically } X_q \approx 0.6\,X_d \]Two-reaction theory splits the armature current into a component \(I_d\) along the direct (pole) axis and a component \(I_q\) along the quadrature axis, each carrying its own reactance:
\[ \mathbf{E}_f = \mathbf{V} + R_a\mathbf{I}_a + jX_d\mathbf{I}_d + jX_q\mathbf{I}_q \]Nothing can be resolved until the q-axis is located, and the trick is that the phasor \(\mathbf{V} + (R_a + jX_q)\mathbf{I}_a\) already lies along it. Its angle is the load angle:
\[ \mathbf{E}'' = \mathbf{V} + (R_a + jX_q)\mathbf{I}_a = E''\angle\delta \]This is the whole of the method. Once \(\delta\) is known the current can be resolved and the rest is arithmetic.
Neglecting \(R_a\), the load angle follows in closed form:
\[ \tan\delta = \frac{I_aX_q\cos\phi}{V \pm I_aX_q\sin\phi} \qquad (+\ \text{generator, lagging};\ -\ \text{leading}) \]With \(R_a\) retained it is easier to work with \(\psi\), the angle between \(\mathbf{I}_a\) and the q-axis:
\[ \tan\psi = \frac{V\sin\phi + I_aX_q}{V\cos\phi + I_aR_a}, \qquad \delta = \psi - \phi \]The current components then follow directly from \(\psi = \phi + \delta\):
\[ I_d = I_a\sin\psi, \qquad I_q = I_a\cos\psi \]The excitation e.m.f. is read along the q-axis, where \(\mathbf{E}_f\) lies by definition. Projecting every phasor onto that axis gives a scalar equation:
\[ E_f = V\cos\delta + I_qR_a + I_dX_d \]The \(X_q\) drop has vanished, because \(jX_q\mathbf{I}_q\) is perpendicular to the q-axis. That is why the method is worth the extra step.
Regulation is defined exactly as for a cylindrical machine once \(E_f\) is known:
\[ \%\,\text{regulation} = \frac{E_f - V}{V}\times100 \]
A 480 V, 60 Hz, delta-connected, four-pole synchronous generator has a direct-axis reactance of 0.1 Ω and a quadrature-axis reactance of 0.075 Ω per phase. Its armature resistance may be neglected. At full load the generator supplies a line current of 1200 A at 0.8 power factor lagging.
- Find the internal generated voltage \(\mathbf{E}_f\) assuming the machine has a cylindrical rotor of reactance \(X_d\).
- Find \(\mathbf{E}_f\) assuming it has a salient-pole rotor, using two-reaction theory.
Get onto a per-phase basis. The winding is in delta, so each phase carries \(1/\sqrt3\) of the line current but stands across the full line voltage:
Delta is the opposite of the star case: the current divides, the voltage does not.
(a) Cylindrical rotor. A uniform air gap means one reactance for the whole current, so the phasor sum is a single term:
(b) Salient-pole rotor: locate the q-axis first. The current cannot be resolved into direct- and quadrature-axis parts until the direction of \(\mathbf{E}_f\) is known, and the standard construction supplies it — add only the quadrature-axis reactance drop:
The magnitude of \(\mathbf{E}''\) is of no interest; only its angle matters, and it gives \(\delta = 4.65^\circ\).
Resolve the armature current. The angle between \(\mathbf{I}_a\) and the q-axis is \(\psi = \phi + \delta\):
As phasors, \(\mathbf{I}_d = 459\angle-85.35^\circ\) (perpendicular to the q-axis) and \(\mathbf{I}_q = 519\angle4.65^\circ\) (along it).
Now assemble \(\mathbf{E}_f\) with each component carrying its own reactance:
The result comes out at exactly the angle predicted by \(\mathbf{E}''\), which is the internal consistency check on the whole construction.
Compare the two models.
| Model | \(E_f\) (V) | \(\delta\) |
|---|---|---|
| Cylindrical rotor, \(X_s = X_d = 0.1\ \Omega\) | 524.5 | 6.07° |
| Salient pole, \(X_d = 0.1,\ X_q = 0.075\ \Omega\) | 524.3 | 4.65° |
The magnitude of \(\mathbf{E}_f\) is barely affected — 0.04 % — but the load angle is considerably different, 4.65° against 6.07°, a change of nearly a quarter.
A 3-phase alternator has a direct-axis synchronous reactance of 0.7 p.u. and a quadrature-axis synchronous reactance of 0.4 p.u. The armature resistance is negligible. At full load and 0.8 power factor lagging determine
- the load angle;
- the no-load per-unit voltage.
Per-unit working makes the data trivial. Rated voltage and rated current are both 1 p.u. by definition of the base:
(a) The load angle, from the closed-form expression that neglects \(R_a\). Note that it is \(X_q\) — not \(X_d\) — that appears, because the q-axis is located by the quadrature reactance drop alone:
Resolve the current onto the two axes. The angle from the q-axis to the current is
(b) The no-load e.m.f., obtained by projecting everything onto the q-axis. With \(R_a = 0\) only two terms survive:
The direct-axis reactance appears here and nowhere else in the calculation; \(X_q\) did its work in locating the axis and then dropped out, because \(jX_q\mathbf{I}_q\) is perpendicular to \(\mathbf{E}_f\).
Check it against the cylindrical estimate. Treating the machine as round with \(X_s = X_d = 0.7\) would give
Within 0.7 % of the two-reaction answer — the same conclusion as Problem 1. The regulation would be 51.5 % either way.
A 3-phase, star-connected, 50 Hz alternator has a direct-axis synchronous reactance of 0.6 p.u. and a quadrature-axis synchronous reactance of 0.45 p.u. The generator delivers rated kVA at rated voltage and 0.8 power factor lagging. The resistive drop at full load is 0.015 p.u. Calculate the load angle, the open-circuit voltage and the voltage regulation.
The per-unit data. Rated kVA at rated voltage means both are unity:
Find \(\psi\) directly. With \(R_a\) retained it is cleaner to compute the angle between the current and the q-axis in one step, rather than \(\delta\) first:
The load angle is then the difference between \(\psi\) and the power-factor angle:
The two current components:
Check: \(I_d^2 + I_q^2 = 0.624 + 0.376 = 1.000\), as it must be for a 1 p.u. current.
Project onto the q-axis to get the open-circuit voltage. Now the resistance term survives, because \(R_a\mathbf{I}_q\) lies along the q-axis:
The resistive term contributes 0.009 of the 1.448 — well under one per cent, yet it shifted \(\psi\) by about half a degree, which is why it was worth keeping in the angle calculation even though it is negligible in the magnitude.
The regulation follows from the same definition used for a cylindrical machine:
A 3-phase, star-connected synchronous generator supplies a current of 10 A at a phase angle of 20° lagging, the terminal voltage being 400 V per phase. The direct-axis reactance is 10 Ω and the quadrature-axis reactance 6.5 Ω per phase; the armature resistance is negligible. Find the load angle, the components \(I_d\) and \(I_q\) of the armature current, the excitation e.m.f. and the voltage regulation.
The given data, per phase. Every reactance is per phase, so the 400 V must be a phase value too — on a star-connected machine that corresponds to \(400\sqrt3 = 693\) V between lines:
Locate the quadrature axis. With \(R_a\) neglected the load angle comes straight from \(X_q\):
Resolve the current about that axis:
The larger component lies along the q-axis, which is what one expects at a fairly high power factor: a nearly in-phase current is mostly torque-producing and only weakly demagnetising.
The direct-axis reactance drop:
The excitation e.m.f., projected onto the q-axis with \(R_a = 0\):
The regulation:
Modest, because at 0.94 power factor the demagnetising component \(I_d\) is less than half the armature current.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Two-reaction phasor equation | \(\mathbf{E}_f = \mathbf{V} + R_a\mathbf{I}_a + jX_d\mathbf{I}_d + jX_q\mathbf{I}_q\) | Per phase — Problem 1 |
| Locating the q-axis | \(\mathbf{E}'' = \mathbf{V} + (R_a+jX_q)\mathbf{I}_a\) | Its angle is \(\delta\); its magnitude is discarded |
| Load angle, \(R_a\) neglected | \(\tan\delta = \dfrac{I_aX_q\cos\phi}{V \pm I_aX_q\sin\phi}\) | \(+\) lagging, \(-\) leading — Problems 2, 4 |
| Angle to the q-axis | \(\tan\psi = \dfrac{V\sin\phi + I_aX_q}{V\cos\phi + I_aR_a}\) | Keeps \(R_a\) — Problem 3 |
| Relation between angles | \(\psi = \phi + \delta\) | Generator, lagging load |
| Direct-axis current | \(I_d = I_a\sin\psi\) | Demagnetising when lagging |
| Quadrature-axis current | \(I_q = I_a\cos\psi\) | Torque-producing component |
| Check on the split | \(I_d^2 + I_q^2 = I_a^2\) | Problem 3 |
| Excitation e.m.f. | \(E_f = V\cos\delta + I_qR_a + I_dX_d\) | \(X_q\) drops out — Problems 2, 3, 4 |
| Cylindrical-rotor limit | \(E_f = \sqrt{(V\cos\phi+I_aR_a)^2+(V\sin\phi+I_aX_s)^2}\) | Valid when \(X_d = X_q\) — Problems 1, 2 |
| Voltage regulation | \(\%\,\text{reg} = (E_f - V)/V \times 100\) | Per phase — Problems 3, 4 |
| Delta connection | \(I_a = I_L/\sqrt3\), \(V = V_L\) | Problem 1 |
| Star connection | \(I_a = I_L\), \(V = V_L/\sqrt3\) | Problem 4 |
| Typical reactance ratio | \(X_q \approx 0.6\,X_d\) | Always \(X_d > X_q\) |
Common Mistakes
Using \(X_d\) to find the load angle. The q-axis is located by the quadrature reactance drop; \(X_d\) enters only afterwards — Problems 2 and 4.
Resolving the current before the axis is known. \(I_d\) and \(I_q\) have no meaning until \(\delta\) has been computed — Problem 1.
Using \(\phi\) instead of \(\psi = \phi + \delta\) when splitting the current. The current is measured from \(\mathbf{V}\) but resolved about the q-axis — Problems 1, 3, 4.
Keeping the \(X_q\) drop in the final scalar equation. \(jX_q\mathbf{I}_q\) is perpendicular to \(\mathbf{E}_f\) and contributes nothing to its magnitude — Problem 2.
Dropping \(R_a\) from the angle calculation because it is small in the magnitude. A 0.015 p.u. resistance shifts \(\psi\) by about half a degree even though it changes \(E_f\) by under 1 % — Problem 3.
Dividing the voltage by \(\sqrt3\) on a delta-connected machine. In delta the phase voltage is the line voltage; it is the current that divides — Problem 1.
Assuming a stated terminal voltage is a line value. Reactances are per phase, so the voltage must be reduced to a phase value if it is not already one — Problem 4.
Concluding that saliency hardly matters because \(E_f\) barely moves. The load angle changes by nearly a quarter, and every stability and power-transfer result depends on \(\delta\) — Problem 1.
Taking \(X_q > X_d\). The direct axis has the shorter air gap and therefore always the larger reactance — every problem here.
Forgetting that \(\delta\) is measured to the q-axis, not to \(\mathbf{E}''\)'s magnitude. \(E'' = 513\) V is not an e.m.f. of the machine; only its direction is used — Problem 1.
Two-reaction theory replaced one reactance with two and, in exchange, asked for one extra construction: locate the quadrature axis before resolving anything. Every problem on this page followed the same four steps — find \(\delta\) from \(X_q\), resolve \(\mathbf{I}_a\) about the axis, project onto it, and read \(E_f\). The reward is a load angle that is right rather than approximately right.
So far the alternator has been considered alone, its terminal voltage decided by its own excitation. Connect it to a network alongside other machines and neither the voltage nor the frequency is its to choose. The load angle then becomes the variable that carries power, and the conditions under which two machines may be joined at all become the first question to answer.
Next: Set 38 — Synchronising and Parallel Operation, where the load angle stops being a by-product of a phasor diagram and becomes the control variable of the machine.